Area between lines

[sigma]

Have a question here that I'm having some trouble on. Heres the question.

\(\displaystyle \
\L\
{\rm Let R be the planar region bounded by the lines }x = 1,x = - 1 \\
{\rm and by }y = x^2 + 5x + 6,{\rm }y = 6 - 5x,{\rm and the x - axis}{\rm . Find the} \\
{\rm area of R}{\rm .} \\)

Is this what the graph should look like?


Setting up the integral:

\(\displaystyle \
\L\
f'(x) = \int_{{\rm - 1}}^{\rm 1} {x^2 + 5x + 6{\rm }dx + \int_{ - 1}^1 {6 - 5x} } {\rm }dx
\\)

Integrating I get:
\(\displaystyle \
\L\
\begin{array}{l}
f(x) = \frac{{x^3 }}{3} + \frac{{5x^2 }}{2} + \left. {6x} \right|_{ - 1}^1 + 6x - \left. {\frac{{5x^2 }}{2}} \right|_{ - 1}^1 \\
= \frac{{\rm 1}}{{\rm 3}} + \frac{5}{2} + 6 + \frac{1}{3} - \frac{5}{2} + 6 + 6 - \frac{5}{2} + 6 - \frac{5}{2} \\
= \frac{{59}}{3} \\
\end{array}
\\)

However, that is not the correct answer. By book got 22/3 as an answer. Where am I going wrong? Am I integrating the wrong area? Help would be greatly appreciated.

 

[pka]

Try this:
\(\displaystyle \
\L\
f'(x) = \int_{{\rm - 1}}^{\rm 0} {x^2 + 5x + 6{\rm }dx + \int_{ 0 }^1 {6 - 5x} } {\rm }dx
\\)

 

[sigma]

Ok. Now I get 22/3. Do you go to zero because the area underneath each line is above or below zero?

 

[tkhunny]

You "go to zero" because x = 0 is where the function intersect.

 

[pka]

Think about how one would describe that region:
“below by the x-axis; from x=-1 to x=0 above by the parabola; and from x=0 to x= by above the line”.

We ‘break’ the integral up to find the corresponding areas.

 

[sigma]

Great. Thanks allot! Really appreciate all this help!